Question

我有以下XML文件：

<RecordSet>
  <Record>
    <ID>001</ID>
    <TermList>
      <Term>Term1</Term>
      <Term>Term2</Term>
      <Term>Term3</Term>
    </TermList>
  </Record>
  <Record>
    <ID>002</ID>
    <TermList>
      <Term>Term3</Term>
      <Term>Term4</Term>
      <Term>Term5</Term>
    </TermList>
  </Record>
</RecordSet>

并且需要将其解析为＆＃34; ID-Term＆＃34;文件，即

001 Term1
001 Term2
001 Term3
002 Term3
002 Term4
002 Term5

目前我有以下申请：

import java.io.IOException;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

import javax.xml.parsers.*;
import javax.xml.xpath.*;

import org.w3c.dom.Document;
import org.w3c.dom.NodeList;
import org.xml.sax.SAXException;

public class MedlineParser {

    public static void main(String[] args) {
        DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
        factory.setNamespaceAware(true);
        DocumentBuilder builder;
        Document doc = null;
        try {
            builder = factory.newDocumentBuilder();
            doc = builder.parse("/home/andrej/Documents/test.xml");
            // Create XPathFactory object
            XPathFactory xpathFactory = XPathFactory.newInstance();
            // Create XPath object
            XPath xpath = xpathFactory.newXPath();
            try {
                XPathExpression expr1 = xpath.compile("/RecordSet/Record/ID/text()");
                NodeList nodes1 = (NodeList) expr1.evaluate(doc, XPathConstants.NODESET);
                for (int i = 0; i < nodes1.getLength(); i++) {
                    String id = nodes1.item(i).getNodeValue();
                    XPathExpression expr2 = xpath.compile("/RecordSet/Record/TermList/Term/text()");
                    NodeList nodes2 = (NodeList) expr2.evaluate(doc, XPathConstants.NODESET);
                    for (int j = 0; j < nodes2.getLength(); j++) {
                        System.out.println(id + " " + nodes2.item(i).getNodeValue());
                    }
                }
            } catch (XPathExpressionException e) {
                e.printStackTrace();
            }

        } catch (IOException | ParserConfigurationException | SAXException e) {
            e.printStackTrace();
        }
    }
}

不幸的是，程序输出目前是：

001 Term1
001 Term1
001 Term1
001 Term1
001 Term1
001 Term1
002 Term2
002 Term2
002 Term2
002 Term2
002 Term2
002 Term2

知道XPath表达式有什么问题吗？

Answer 1

两个问题：

XPath必须将第一个循环中迭代的Term节点的索引纳入帐户。您的当前XPath每次为每个ID节点获取所有 XPathExpression expr2 = xpath.compile("/RecordSet/Record[" + (i + 1) + "]/TermList/Term/text()");个节点。您应该将其更改为：
```
for
```
内部j循环中有拼写错误。您应该使用i代替for (int j = 0; j < nodes2.getLength(); j++) { System.out.println(id + " " + nodes2.item(j).getNodeValue()); }：
```
df1
```

Answer 2

似乎您正在打印所有ID和术语的笛卡尔积。

这会更容易：

使用XPath表达式/RecordSet/Record选择并循环遍历所有Record节点。
对于每个记录节点，使用Record-node作为上下文节点，选择id（使用XPath ID）和术语（使用XPath Termlist/Term）。

XPath选择Java中的节点列表

2 个答案: